I noticed that the coefficients 89 and 144 are Fibonacci numbers. They are the 11th and 12th, and the exponents are also 11 and 12. So this suggests phi is a root, or equivalently x^2-x-1 is a factor.
So lets expand all the coefficients, even the zero coefficients. Then we get this lengthy polynomial expression:
89x^12 + (-89-55)x^11 + (-89+55+34)x^10 + (55-34-21)x^9 + (-34+21+13)x^8 + (21-13-8)x^7 + (-13+8+5)x^6 + (8-5-3)x^5 + (-5+3+2)x^4 + (3-2-1)x^3 + (-2+1+1)x^2 + (1-1)x - 1 = 0
So now collect all the common coefficients:
89(x^12-x^11-x^10) - 55(x^11-x^10-x^9) + 34(x^10-x^9-x^8) - 21(x^9-x^8-x^7) + 13(x^8-x^7-x^6) - 8(x^7-x^6-x^5) + 5(x^6-x^5-x^4) - 3(x^5-x^4-x^3) + 2(x^4-x^3-x^2) - 1(x^3-x^2-x) + 1(x^2-x-1) = 0
Now it is easy to see that x^2-x-1 is a factor. The whole polynomial factors into:
(x^2-x-1) * (89x^10-55x^9+34x^8-21x^7+13x^6-8x^5+5x^4-3x^3+2x^2-x+1) = 0.
So then the two roots of x^2-x-1 are phi and -1/phi, which are then also roots of the original equation.
Solution
